CHAPTER5_SECTION8 - Chapter 5 The First Law of...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Chapter 5 The First Law of Thermodynamics 5-221 Two rigid tanks that contain water at different states are connected by a valve. The valve is opened and the two tanks come to the same state at the temperature of the surroundings. The final pressure and the amount of heat transfer are to be determined. Assumptions 1 The tanks are stationary and thus the kinetic and potential energy changes are zero. 2 The tank is insulated and thus heat transfer is negligible. 3 There are no work interactions. Analysis We take the entire contents of the tank as the system. This is a closed system since no mass enters or leaves. Noting that the volume of the system is constant and thus there is no boundary work, the energy balance for this stationary closed system can be expressed as ] ) ( ) ( [ ] [ 0) = PE = KE (since ) ( ) ( 1 1 1 1 2 total 2, , 1 , 1 , 2 out out energies etc. potential, kinetic, internal, in Change system mass and work, heat, by nsfer energy tra Net out in B A B A B A B A u m u m u m U U U Q W U U U Q E E E        The properties of water in each tank are (Tables A-4 through A-6) Tank A: kJ/kg 1949.3 , 31 . 604 /kg m 0.4625 , 001084 . 0 80 . 0 kPa 400 3 1 1 fg f g f u u v v x P kJ/kg 2163.75 3 . 1949 8 . 0 31 . 604 /kg m 0.3702 001084 . 0 4625 . 0 8 . 0 001084 . 0 1 , 1 3 1 , 1 fg f A fg f A u x u u v x v v Tank B: kJ/kg 2304.9 , 88 . 104 /kg m 43.36 , 001003 . 0 /kg m 0.731 C 25 /kg m 0.731 kg 0.957 m 0.7 kg 0.957 417 . 0 540 . 0 kg 0.417 /kg m 1.1988 m 0.5 kg 0.540 /kg m 0.3702 m 0.2 kJ/kg 2731.2 /kg m 1.1988 C 250 kPa 200 3 3 2 2 3 3 2 , 1 , 1 3 3 , 1 , 1 3 3 , 1 , 1 , 1 3 , 1 1 1 fg f g f t t B A t B B B A A A B B u u v v v T m V v m m m v V m v V m u v T P Thus at the final state the system will be a saturated liquid-vapor mixture since v f < v 2 < v g . Then the final pressure must be P 2 = P sat @ 25 C = 3.169 kPa Also, kJ/kg 143.60 2304.9 0.0168 104.88 0168 . 0 001 . 0 36 . 43 001 . 0 731 . 0 2 2 2 2 fg f fg f u x u u v v v x Substituting, Q out = -[(0.957)(143.6) - (0.540)(2163.75) - (0.417)(2731.2)] = 2170 kJ 5-208 H 2 O 400 kPa H 2 O 200 kPa B A Q
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Chapter 5 The First Law of Thermodynamics 5-222 Problem 5-221 is reconsidered. The effect of the environment temperature on the final pressure and the heat transfer as the environment temperature varies from 0°C to 50°C is to be investigated. The final results are to be plotted against the environment temperature. "Knowns" Vol_A=0.2 "[m^3]" P_A[1]=400 "[kPa]" x_A[1]=0.8 T_B[1]=250 "[C]" P_B[1]=200 "[kPa]" Vol_B=0.5 "[m^3]" T_final=25 "[C]" "T_final = T_surroundings. To do the parametric study or to solve the problem when Q_out = 0, place this statement in {}." {Q_out=0"[kJ]"} "To determine the surroundings temperature that makes Q_out = 0, remove the {} and resolve the problem." "Solution" "Conservation of Energy for the combined tanks:" E_in-E_out=DELTAE E_in=0 "[kJ]" E_out=Q_out "[kJ]" DELTAE=m_A*(u_A[2]-u_A[1])+m_B*(u_B[2]-u_B[1]) "[kJ]" m_A=Vol_A/v_A[1] "[kg]" m_B=Vol_B/v_B[1] "[kg]" u_A[1]=INTENERGY(Steam,P=P_A[1], x=x_A[1]) "[kJ/kg]" v_A[1]=volume(Steam,P=P_A[1], x=x_A[1])
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern